I am attempting to calculate the vertical carrier mobility through a semiconductor powder sample coated on FTO using Spin Coating method. The I-V results I recorded using SMU 2401, shows current saturation at very low voltage like 0.3V, 0.4V etc. From Literature I found the saturation Value to be more than 2V. What could be the reason of showing this much low current saturation in my sample. I want know the possible reasons behind such result. whether the device is short? or any other reasons? I am attaching the images of (a) the configuration I chose for coating and I-V measurements (b) Corresponding I-V Graph of my sample (c) reference I-V plot taken from literature .
It looks like the device is short. You did not mention how you fabricated the device. So, I can't help you much other than a few simple ideas as you can go with thicker Cu2O film first to check if it still gives a similar result or not. otherwise, you can try to deposit the top metal layer with the E-beam technique.
Karimul Islam, I sonicated the Cu2O powder sample in a PMMA solution and then spin-coated it over FTO. I coated the top electrode silver, manually.
Change the compliance current in the instrument. It seems like you have set the compliance current to 10uA. Set it to 100mA.
Dear Unniyarcha K.K
Cu2O syntheses in three different crystals morphology: octahedron is highly conductive, a cube is moderately conductive, and a rhombic dodecahedron is nonconductive. The conductivity differences are ascribed to the presence of a thin surface layer having different degrees of band bending [1]. In this respect, each type has its syntheses route: Octahedral Cu2O microcrystals obtain in the absence of oleylamine and at the highest sodium oleate concentration. Rhombic dodecahedral Cu2O microcrystals obtain at the highest oleylamine concentration or at the lowest sodium oleate concentration[2].
[1]Article Facet-Dependent Electrical Conductivity Properties of Cu2O Crystals
[2]Article Morphology‐controlled Synthesis of Octahedral‐to‐Rhombic Dod...
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